$f\,^{\prime}(x)=4e^x$ and $f(2)=16+4e^2$. $f(0) = $
Finding $f(x)$ We have $f'(x)=4e^x$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (4e^x)\,dx \\\\ & = {4e^x} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(2)=16+4e^2$. Here's what we get when we plug in $2$ : $\begin{aligned}f(2)&={4e^{2}} {+ C} \end{aligned}$ We are given that this must equal $16+4e^2$ : $16+4e^2 = {4e^2} {+ C}$ Solving the equation gives us ${C=16}$. Finding $f(0)$ Now, we have that $f(x)={4e^x} {+ 16}$. Let's find $f(0)$ by plugging in $0$ : $\begin{aligned}f(0)&=4e^0 + 16\\\\ &=20 \end{aligned}$ The answer $f(0) = 20$